Does every path connected subspace of $\mathbb{R}^2$ have a fundamental group the trivial or an infinity group?
For example, for convex subspaces we know that, but if we take only path connected subespaces what happen?
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Every reasonable path connected subspace of $\mathbb{R}^2$ has fundamental group a free group, which is either trivial or infinite. "Reasonable" means homotopy equivalent to a tubular neighborhood of it in $\mathbb{R}^2$, and in particular means homotopy equivalent to an open subset of $\mathbb{R}^2$. Now, an open subset of $\mathbb{R}^2$ is a noncompact surface, and every connected noncompact surface has fundamental group a free group; see, for example, this MO question.
An example of an unreasonable path connected subspace of $\mathbb{R}^2$ is the Hawaiian earring, whose fundamental group is not free.
Qiaochu Yuan
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1However, Hawaiian earring has an infinite fundamental group – sinbadh Apr 25 '16 at 04:15
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Do you know any easy argument to prove that fundamental group of hawiian earring is not free?? I know that it is uncountable, but proving it is not free seems difficult. – Anubhav Mukherjee Apr 25 '16 at 18:30
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Your question was answered in affirmative by George Lowther here:
Is the fundamental group of every subset of $\mathbb{R}^2$ torsion free?
More precisely, he proves that if $X$ is a subset of any closed surface $S$, for instance, $S=S^2$, and $x\in X$, then $\pi_1(X,x)$ is torsion free.
Note that no assumption is made about the nature of the subset $X$, it is not assumed to be "reasonable".
From this, it follows that if $\pi_1(X,x)$ is nontrivial, it contains a nontrivial element $\gamma$, which, then, necessarily has infinite order. Therefore, $\pi_1(X,x)$ is either trivial or infinite.
Moishe Kohan
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